Project no 0001 Date 11/09/2019 Notes Revision File Sample ... › wp-content › uploads › ...For tube and fitting scaffolding, in accordance with BS EN 12811-1:2003 and NASC TG20:13

Project no 0001 Date 11/09/2019Name Sample project Prepared by TRItem Scaffold 001 Checked by BBNotes RevisionFile Sample brick guard scaffold.ssc Page 1 of 14

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SMART Calculations version 19.0.31. Copyright © Computer and Design Services Ltd. 2019

Tied independent scaffolding leg loadsFor tube and fitting scaffolding, in accordance with BS EN 12811-1:2003 and NASC TG20:13.

ℹ This calculation should be read in conjunction with the wind factor and tie duty calculation reports.

Site locationDescription Value

Site address East Overcliff Drive, E Overcliff Dr, Bournemouth BH1, UK

TG20:13 wind factor, STG20:13 26.6

Peak velocity pressure at 13.00 m, qp(z = 13.00m) 0.888 kN/m²




Scaffold dimensionsDescription Value

Number of boarded lifts, nb 2

Number of unboarded lifts, nu 4

Maximum lift height, Hlift 2.00 m

Maximum bay length, Lbay 2.00 m

Number of main boards wide, nm 5

Number of inside boards, ni 2

Edge protectionDescription Value

Guard rails at boarded lifts, ngr,b 2

Guard rails at unboarded lifts, ngr,u 1

Inner guard rails at boarded lifts None

Inner guard rails at unboarded lifts None

Inner toe boards None

Scaffold configurationDescription Value

Cladding Brick guards

Facade permeability (1) Impermeable

Tie pattern TG20:13 A

Structural transoms None

(1) No significant openings.

LoadingDescription Value

Main platform working load, Pm 2.00 kN/m²

Inner platform working load, Pi 0.75 kN/m²

Number of loaded lifts, nl 1

Number of 50% loaded lifts, nl,50 1

Component dimensionsDescription Value

Tube diameter, dt 48.3 mm

Board width, Wb 225 mm

Board thickness, tb 38 mm

Component weightsComponent Unit mass Unit weight

Tubes, Pt 4.37 kg/m 0.043 kN/m

Boards, Pb 25.00 kg/m² 0.245 kN/m²

Right-angle couplers, Pra 1.25 kg 0.012 kN

Swivel couplers, Psc 1.33 kg 0.013 kN

Putlog couplers, Ppc 1.00 kg 0.010 kN

Brick guards, Pbg 1.60 kg/m² 0.016 kN/m²

Calculated dimensionsToe board thickness ttb = 0.038 m

Main platform width between ledger centres

Wm = nm ⋅ Wb + ttb - dt = 5 ⋅ 0.225 + 0.038 - 0.048 = 1.115 m

Cantilever width beyond inner ledger

Wi = ni ⋅ Wb + 1.5 ⋅ dt = 2 ⋅ 0.225 + 1.5 ⋅ 0.048 = 0.522 m

Total platform width Wtotal = Wm + Wi = 1.115 + 0.522 = 1.637 m




Transverse tube oversail length Lt,tr = 0.100 m

Service gap Wgap = 0.15 m

Board transom or end guard rail length

Ltr = Wtotal + 2 ⋅ Lt,tr = 1.637 + 2 ⋅ 0.100 = 1.837 m

Tie tube length Ltt = Wtotal + Lt,tr + Wgap = 1.637 + 0.100 + 0.15 = 1.887 m

Bracing tube oversail length Lt,b = 0.200 m

Facade brace analytical model length

Lfb,1 = Lbay2 + Hlift2 = 2.002 + 2.002 = 2.828 m

Facade brace length Lfb = Lfb,1 + 2 ⋅ Lt,b = 2.828 + 2 ⋅ 0.200 = 3.228 mLedger brace length Llb = Wm2 + Hlift2 + 2 ⋅ Lt,b = 1.1152 + 2.002 + 2 ⋅ 0.200 = 2.690 m

Maximum board transoms per bay nt = ceiling(Lbay1.2) + 1 = ceiling(2.001.2 ) + 1 = 3

Dead loadingDead load per standard per lift Fst = Hlift ⋅ Pt = 2.00 ⋅ 0.043 = 0.086 kNDead load per ledger brace Flb = Llb ⋅ Pt = 2.690 ⋅ 0.043 = 0.115 kNDead load of a board transom or end guard rail

Ftr = Pt ⋅ Ltr = 0.043 ⋅ 1.837 = 0.079 kN

Dead load of a tie tube Ftt = Pt ⋅ Ltt = 0.043 ⋅ 1.887 = 0.081 kNDead load of toe boards Ftb = Pb ⋅ Wb = 0.245 ⋅ 0.225 = 0.055 kN/mDead load of an end toe board Fetb = Ftb ⋅ Ltr = 0.055 ⋅ 1.837 = 0.101 kNDead load of brick guards Fbg = Pbg ⋅ Hbg = 0.016 ⋅ 1.0 = 0.016 kN/mDead load of the end brick guards

Febg = Wtotal ⋅ Fbg = 1.637 ⋅ 0.016 = 0.026 kN

Dead loading on the outer ledgersDistribution factor for permanent dead loading to the outer face

fo =0.5 ⋅ (Wm - Wi)

Wm=

0.5 ⋅ (1.115 - 0.522)

1.115= 0.266

Dead load of board transoms at the outer face Ftr,o = (fo ⋅ Ftr + Ppc) ⋅

ntLbay

= (0.266 ⋅ 0.079 + 0.010) ⋅3

2.00= 0.046 kN/m

Dead load of boards at the outer face

Fb,o = 0.5 ⋅ Wm ⋅ Pb = 0.5 ⋅ 1.115 ⋅ 0.245 = 0.137 kN/m

Dead load on unboarded lift outer ledgers

Fl,u,o = Pt + Ftr,o = 0.043 + 0.046 = 0.089 kN/m

Dead load on boarded lift outer ledgers

Fl,b,o = Pt + Ftr,o + Ftb + Fb,o + Fbg = 0.043 + 0.046 + 0.055 + 0.137 + 0.016= 0.297 kN/m




Dead loading on the outer standardsDead load per tie tube at the outer face

Ftt,o = fo ⋅ Ftt + Pra = 0.266 ⋅ 0.081 + 0.012 = 0.034 kN

Dead load per unbraced standard per lift

Fs,u = Fst = 0.086 kN

Dead load per ledger-braced standard per lift

Fs,l = Fst + 0.5 ⋅ Flb + Psc = 0.086 + 0.5 ⋅ 0.115 + 0.013 = 0.156 kN

Dead load of couplers to outer standards at unboarded lifts

Fc,u,o = (1 + ngr,u) ⋅ Pra = (1 + 1) ⋅ 0.012 = 0.025 kN

Dead load of couplers to outer standards at boarded lifts

Fc,b,o = (1 + ngr,b) ⋅ Pra + Ppc = (1 + 2) ⋅ 0.012 + 0.010 = 0.047 kN

Dead load per outer end standard per unboarded lift Fes,u,o = Fs,l + fo ⋅ ngr,u ⋅ Ftr + ngr,u ⋅ Pt ⋅

Lbay2

+ 2 ⋅ Fc,u,o

= 0.156 + 0.266 ⋅ 1 ⋅ 0.079 + 1 ⋅ 0.043 ⋅2.00

2+ 2 ⋅ 0.025 = 0.269 kN

Dead load per outer end standard per boarded lift Fes,b,o = Fs,l + fo ⋅ (ngr,b ⋅ Ftr + Fetb + Febg) + ngr,b ⋅ Pt ⋅

Lbay2

+ 2 ⋅ Fc,b,o

= 0.156 + 0.266 ⋅ (2 ⋅ 0.079 + 0.101 + 0.026) + 2 ⋅ 0.043 ⋅2.00

2+ 2 ⋅ 0.047

= 0.411 kN

Dead load per outer unbraced intermediate standard per unboarded lift

Fis,u,u,o = ngr,u ⋅ Pt ⋅ Lbay + Fs,u + Fc,u,o = 1 ⋅ 0.043 ⋅ 2.00 + 0.086 + 0.025 = 0.196 kN

Dead load per outer ledger-braced intermediate standard per unboarded lift

Fis,u,b,o = ngr,u ⋅ Pt ⋅ Lbay + Fs,l + Fc,u,o = 1 ⋅ 0.043 ⋅ 2.00 + 0.156 + 0.025 = 0.267 kN

Dead load per outer unbraced intermediate standard per boarded lift

Fis,b,u,o = ngr,b ⋅ Pt ⋅ Lbay + Fs,u + Fc,b,o = 2 ⋅ 0.043 ⋅ 2.00 + 0.086 + 0.047 = 0.304 kN

Dead load per outer ledger-braced intermediate standard per boarded lift

Fis,b,b,o = ngr,b ⋅ Pt ⋅ Lbay + Fs,l + Fc,b,o = 2 ⋅ 0.043 ⋅ 2.00 + 0.156 + 0.047 = 0.375 kN

Dead loading on the inner ledgersDistribution factor for dead loading to the inner face fi =

0.5 ⋅ (Wm + Wi)

Wm=

0.5 ⋅ (1.115 + 0.522)

1.115= 0.734

Dead load of board transoms at the inner face Ftr,i = (fi ⋅ Ftr + Ppc) ⋅

ntLbay

= (0.734 ⋅ 0.079 + 0.010) ⋅3

2.00= 0.101 kN/m

Dead load of boards at the inner face

Fb,i = fi ⋅ Wtotal ⋅ Pb = 0.734 ⋅ 1.637 ⋅ 0.245 = 0.295 kN/m

Dead load on unboarded lift inner ledgers

Fl,u,i = Pt + Ftr,i = 0.043 + 0.101 = 0.144 kN/m

Dead load on boarded lift inner ledgers

Fl,b,i = Pt + Ftr,i + Fb,i = 0.043 + 0.101 + 0.295 = 0.439 kN/m




Dead loading on the inner standardsDead load per tie tube at the inner face

Ftt,i = fi ⋅ Ftt + Pra = 0.734 ⋅ 0.081 + 0.012 = 0.072 kN

Dead load of couplers to inner standards at unboarded lifts

Fc,u,i = (1 + ngr,u,i) ⋅ Pra = (1 + 0) ⋅ 0.012 = 0.012 kN

Dead load of couplers to inner standards at boarded lifts

Fc,b,i = (1 + ngr,b,i) ⋅ Pra = (1 + 0) ⋅ 0.012 = 0.012 kN

Dead load per inner end standard per unboarded lift Fes,u,i = Fs,l + fi ⋅ ngr,u ⋅ Ftr + ngr,u,i ⋅ Pt ⋅

Lbay2

+ Fc,u,o + Fc,u,i

= 0.156 + 0.734 ⋅ 1 ⋅ 0.079 + 0 ⋅ 0.043 ⋅2.00

2+ 0.025 + 0.012 = 0.251 kN

Dead load per inner end standard per boarded lift Fes,b,i = Fs,l + fi ⋅ (ngr,b ⋅ Ftr + Fetb + Febg) + ngr,b,i ⋅ Pt ⋅

Lbay2

+ Fc,b,o + Fc,b,i

= 0.156 + 0.734 ⋅ (2 ⋅ 0.079 + 0.101 + 0.026) + 0 ⋅ 0.043 ⋅2.00

2+ 0.047 + 0.012

= 0.424 kN

Dead load per inner unbraced intermediate standard per unboarded lift

Fis,u,u,i = ngr,u,i ⋅ Pt ⋅ Lbay + Fs,u + Fc,u,i = 0 ⋅ 0.043 ⋅ 2.00 + 0.086 + 0.012 = 0.098 kN

Dead load per inner ledger-braced intermediate standard per unboarded lift

Fis,u,b,i = ngr,u,i ⋅ Pt ⋅ Lbay + Fs,l + Fc,u,i = 0 ⋅ 0.043 ⋅ 2.00 + 0.156 + 0.012 = 0.169 kN

Dead load per inner unbraced intermediate standard per boarded lift

Fis,b,u,i = ngr,b,i ⋅ Pt ⋅ Lbay + Fs,u + Fc,b,i = 0 ⋅ 0.043 ⋅ 2.00 + 0.086 + 0.012 = 0.098 kN

Dead load per inner ledger-braced intermediate standard per boarded lift

Fis,b,b,i = ngr,b,i ⋅ Pt ⋅ Lbay + Fs,l + Fc,b,i = 0 ⋅ 0.043 ⋅ 2.00 + 0.156 + 0.012 = 0.169 kN

Dead loading of the facade bracingDead load of facade bracing

Ffb =Pt ⋅ Lfb + 2 ⋅ Psc

Lfb,1=

0.043 ⋅ 3.228 + 2 ⋅ 0.0132.828

= 0.058 kN/m

Imposed loadingImposed load per loaded lift at the outer face

Fi,o = 0.5 ⋅ Pm ⋅ Wm = 0.5 ⋅ 2.00 ⋅ 1.115 = 1.115 kN/m

Imposed load per loaded lift at the inner face Fi,i = 0.5 ⋅ Pm ⋅ Wm + Pi ⋅ Wi +

0.5 ⋅ Pi ⋅ Wi2

Wm

= 0.5 ⋅ 2.00 ⋅ 1.115 + 0.75 ⋅ 0.522 +0.5 ⋅ 0.75 ⋅ 0.5222

1.115= 1.598 kN/m

Imposed load per 50% loaded lift at the outer face

Fi,o,50 = 0.5 ⋅ Fi,o = 0.5 ⋅ 1.115 = 0.557 kN/m

Imposed load per 50% loaded lift at the inner face

Fi,i,50 = 0.5 ⋅ Fi,i = 0.5 ⋅ 1.598 = 0.799 kN/m




Out-of-service imposed loadingOut-of-service imposed load on the main platform (EN 12811-1 cl.6.2.9.2.b2)

Pm,o = 0.50 kN/m²

Out-of-service imposed load on the inner platform

Pi,o = 0.00 kN/m²

Out-of-service imposed load per loaded lift at the outer face

Fio,o = 0.5 ⋅ Pm,o ⋅ Wm = 0.5 ⋅ 0.50 ⋅ 1.115 = 0.279 kN/m

Out-of-service imposed load per loaded lift at the inner face

Fio,i = 0.5 ⋅ Pm,o ⋅ Wm = 0.5 ⋅ 0.50 ⋅ 1.115 = 0.279 kN/m

Wind coefficientsDescription Value

Site coefficient for parallel wind on unclad scaffolding, cs||u

1.0

Aerodynamic force coefficient for tubes, cf,t 1.2

Aerodynamic force coefficient for board-bearing transom tubes, cf,tr

0.8

Aerodynamic force coefficient (normal) for toe boards, cf,b

1.3

Description Value

Aerodynamic force coefficient (parallel) for toe boards, cf||tb

0.1

Aerodynamic force coefficient (parallel) for platform boards, cf||b

0.02

Aerodynamic force coefficient (normal) for brick guards, cf,bg

0.177

Aerodynamic force coefficient (parallel) for brick guards, cf||bg

0.073

In-service wind loadingIn-service wind peak velocity pressure

qp,i = 0.200 kN/m²

In-service wind loading on the end framesReference height for materials at in-service working lifts, TG20:13 Design Guide Table 4.3

Hr,m = 0.438 m

Toe board height Htb = 0.225 m

Tube length of boarded end frame

Lt,be = 2 ⋅ Hlift + Llb + ngr,b ⋅ Ltr = 2 ⋅ 2.00 + 2.690 + 2 ⋅ 1.837 = 10.364 m

Tube length of unboarded end frame

Lt,ue = 2 ⋅ Hlift + Llb + ngr,u ⋅ Ltr = 2 ⋅ 2.00 + 2.690 + 1 ⋅ 1.837 = 8.527 m

Area of toe boards and materials at in-service working lift end frame

Atb,we = Hr,m ⋅ Wm + (Htb + tb) ⋅ Wi = 0.438 ⋅ 1.115 + (0.225 + 0.038) ⋅ 0.522= 0.626 m²

In-service wind load on working lift end frame

Fwi||we = qp,i ⋅ cs||u ⋅ (cf,t ⋅ Lt,be ⋅ dt + cf,b ⋅ Atb,we)= 0.200 ⋅ 1.0 ⋅ (1.2 ⋅ 10.364 ⋅ 0.048 + 1.3 ⋅ 0.626) = 0.283 kN




In-service wind load on unboarded lift end frame

Fwi||ue = qp,i ⋅ cs||u ⋅ cf,t ⋅ Lt,ue ⋅ dt = 0.200 ⋅ 1.0 ⋅ 1.2 ⋅ 8.527 ⋅ 0.048 = 0.099 kN

Area of brick guards at working lift end frame

Abg,we = (1.0 - Hr,m) ⋅ Wtotal = (1.0 - 0.438) ⋅ 1.637 = 0.920 m²

In-service wind load on working lift end frame with brick guards

Fwi||bgwe = qp,i ⋅ cs||u ⋅ cf,bg ⋅ Abg,we + Fwi||we = 0.200 ⋅ 1.0 ⋅ 0.177 ⋅ 0.920 + 0.283= 0.315 kN

In-service wind loading per bayTube length of unbraced intermediate frame

Lt,ui = 2 ⋅ Hlift = 2 ⋅ 2.00 = 4.000 m

Tube length of ledger-braced intermediate frame

Lt,bi = 2 ⋅ Hlift + Llb = 2 ⋅ 2.00 + 2.690 = 6.690 m

Tube length of board transoms per bay

Ltr = nt ⋅ Ltr = 3 ⋅ 1.837 = 5.511 m

In-service parallel wind on boards and transoms per bay

Fwi||b = qp,i ⋅ cs||u ⋅ (cf||b ⋅ Wtotal ⋅ Lbay + cf,tr ⋅ Ltr ⋅ dt)= 0.200 ⋅ 1.0 ⋅ (0.02 ⋅ 1.637 ⋅ 2.00 + 0.8 ⋅ 5.511 ⋅ 0.048) = 0.056 kN

In-service parallel wind on toe boards per bay

Fwi||tb = qp,i ⋅ cs||u ⋅ cf||tb ⋅ Htb ⋅ Lbay = 0.200 ⋅ 1.0 ⋅ 0.1 ⋅ 0.225 ⋅ 2.00 = 0.009 kN

In-service parallel wind on brick guards per bay

Fwi||bg = qp,i ⋅ cs||u ⋅ cf||bg ⋅ Lbay ⋅ 1.0 = 0.200 ⋅ 1.0 ⋅ 0.073 ⋅ 2.00 ⋅ 1.0 = 0.029 kN

In-service wind on transoms per unboarded lift bay

Fwi||tu = qp,i ⋅ cs||u ⋅ cf,t ⋅ Ltr ⋅ dt = 0.200 ⋅ 1.0 ⋅ 1.2 ⋅ 5.511 ⋅ 0.048 = 0.064 kN

In-service wind on unbraced intermediate frame tubes

Fwi||ui = qp,i ⋅ cs||u ⋅ cf,t ⋅ Lt,ui ⋅ dt = 0.200 ⋅ 1.0 ⋅ 1.2 ⋅ 4.000 ⋅ 0.048 = 0.046 kN

In-service wind on ledger-braced intermediate frame tubes

Fwi||bi = qp,i ⋅ cs||u ⋅ cf,t ⋅ Lt,bi ⋅ dt = 0.200 ⋅ 1.0 ⋅ 1.2 ⋅ 6.690 ⋅ 0.048 = 0.078 kN

In-service wind loading distribution to the windward end standardsWind load distribution factor to outer standards fo =

0.5 ⋅ (Wm - Wi)

Wm=

0.5 ⋅ (1.115 - 0.522)

1.115= 0.266

Wind load distribution factor to inner standards fi =

0.5 ⋅ (Wm + Wi)

Wm=

0.5 ⋅ (1.115 + 0.522)

1.115= 0.734

In-service wind on working lift outer end standards

Fwi||we,o = fo ⋅ Fwi||bgwe = 0.266 ⋅ 0.315 = 0.084 kN

In-service wind on working lift inner end standards

Fwi||we,i = fi ⋅ Fwi||bgwe = 0.734 ⋅ 0.315 = 0.232 kN

In-service wind on unboarded lift outer end standards

Fwi||ue,o = fo ⋅ Fwi||ue = 0.266 ⋅ 0.099 = 0.026 kN

In-service wind on unboarded lift inner end standards

Fwi||ue,i = fi ⋅ Fwi||ue = 0.734 ⋅ 0.099 = 0.073 kN




In-service wind loading distribution to the intermediate standardsIn-service wind on unbraced intermediate outer standards of boarded lifts

Fwi||ui,b,o = 0.5 ⋅ Fwi||ui + fo ⋅ Fwi||b + Fwi||tb + Fwi||bg= 0.5 ⋅ 0.046 + 0.266 ⋅ 0.056 + 0.009 + 0.029 = 0.076 kN

In-service wind on unbraced intermediate inner standards of boarded lifts

Fwi||ui,b,i = 0.5 ⋅ Fwi||ui + fi ⋅ Fwi||b = 0.5 ⋅ 0.046 + 0.734 ⋅ 0.056 = 0.064 kN

In-service wind on ledger-braced outer intermediate standards of boarded lifts

Fwi||bi,b,o = 0.5 ⋅ Fwi||bi + fo ⋅ Fwi||b + Fwi||tb + Fwi||bg= 0.5 ⋅ 0.078 + 0.266 ⋅ 0.056 + 0.009 + 0.029 = 0.092 kN

In-service wind on ledger-braced inner intermediate standards of boarded lifts

Fwi||bi,b,i = 0.5 ⋅ Fwi||bi + fi ⋅ Fwi||b = 0.5 ⋅ 0.078 + 0.734 ⋅ 0.056 = 0.080 kN

In-service wind on unbraced intermediate outer standards of unboarded lifts

Fwi||ui,u,o = 0.5 ⋅ Fwi||ui,u + fo ⋅ Fwi||tu = 0.5 ⋅ 0.046 + 0.266 ⋅ 0.064 = 0.040 kN

In-service wind on unbraced intermediate inner standards of unboarded lifts

Fwi||ui,u,i = 0.5 ⋅ Fwi||ui,u + fi ⋅ Fwi||tu = 0.5 ⋅ 0.046 + 0.734 ⋅ 0.064 = 0.070 kN

In-service wind on ledger-braced intermediate outer standards of unboarded lifts

Fwi||bi,u,o = 0.5 ⋅ Fwi||bi,u + fo ⋅ Fwi||tu = 0.5 ⋅ 0.078 + 0.266 ⋅ 0.064 = 0.056 kN

In-service wind on ledger-braced intermediate inner standards of unboarded lifts

Fwi||bi,u,i = 0.5 ⋅ Fwi||bi,u + fi ⋅ Fwi||tu = 0.5 ⋅ 0.078 + 0.734 ⋅ 0.064 = 0.086 kN

In-service wind distribution to the leeward end standardsIn-service wind on working lift outer leeward end standards

Fwi||we,l,o = fo ⋅ Fwi||bgwe + fo ⋅ Fwi||b + Fwi||tb + Fwi||bg= 0.266 ⋅ 0.315 + 0.266 ⋅ 0.056 + 0.009 + 0.029 = 0.137 kN

In-service wind on working lift inner leeward end standards

Fwi||we,l,i = fi ⋅ Fwi||bgwe + fi ⋅ Fwi||b = 0.734 ⋅ 0.315 + 0.734 ⋅ 0.056 = 0.272 kN

In-service wind on unboarded lift outer leeward end standards

Fwi||ue,l,o = fo ⋅ Fwi||ue + fo ⋅ Fwi||tu = 0.266 ⋅ 0.099 + 0.266 ⋅ 0.064 = 0.043 kN

In-service wind on unboarded lift inner leeward end standards

Fwi||ue,l,i = fi ⋅ Fwi||ue + fi ⋅ Fwi||tu = 0.734 ⋅ 0.099 + 0.734 ⋅ 0.064 = 0.120 kN

In-service wind loading on additional tubesIn-service wind on facade bracing Fwi||fb = qp,i ⋅ cs||u ⋅ cf,t ⋅ dt ⋅

HliftLfb,1

= 0.200 ⋅ 1.0 ⋅ 1.2 ⋅ 0.048 ⋅2.00

2.828= 0.008 kN/m

In-service wind per tie tube at the inner face

Fwi||tt,i = fi ⋅ qp,i ⋅ cs||u ⋅ cf,t ⋅ dt ⋅ Ltt = 0.734 ⋅ 0.200 ⋅ 1.0 ⋅ 1.2 ⋅ 0.048 ⋅ 1.887= 0.016 kN

In-service wind per tie tube at the outer face

Fwi||tt,o = fo ⋅ qp,i ⋅ cs||u ⋅ cf,t ⋅ dt ⋅ Ltt = 0.266 ⋅ 0.200 ⋅ 1.0 ⋅ 1.2 ⋅ 0.048 ⋅ 1.887= 0.006 kN




Out-of-service wind loadingThe out-of-service peak velocity pressure ( ) varies with the height of the scaffold in accordance with clauses 5.3, 7.6 qp z(2), 7.9.2 (5) and 7.11 (4) of BS EN 1991-1-4:2005 + A1:2010.

Out-of-service peak velocity pressure at 2.00 m

qp(z = 2.00m) = 0.713 kN/m²

Out-of-service peak velocity pressure at 13.00 m

qp(z = 13.00m) = 0.888 kN/m²

Out-of-service wind loading on the end framesArea of toe boards at non-working boarded lift end frame

Atb,ne = Wtotal ⋅ (Htb + tb) = 1.637 ⋅ (0.225 + 0.038) = 0.431 m²

Out-of-service wind load on boarded lift end frame

Fwo||be = qp(z = 13.00m) ⋅ cs||u ⋅ (cf,t ⋅ Lt,be ⋅ dt + cf,b ⋅ Atb,ne)= 0.888 ⋅ 1.0 ⋅ (1.2 ⋅ 10.364 ⋅ 0.048 + 1.3 ⋅ 0.431) = 1.030 kN

Out-of-service wind load on unboarded lift end frame

Fwo||ue = qp(z = 8.00m) ⋅ cs||u ⋅ cf,t ⋅ Lt,ue ⋅ dt = 0.838 ⋅ 1.0 ⋅ 1.2 ⋅ 8.527 ⋅ 0.048= 0.414 kN

Out-of-service wind load on boarded lift end frame with brick guards

Fwo||bgne = qp(z = 13.00m) ⋅ cs||u ⋅ cf,bg ⋅ Abg,ne + Fwo||be= 0.888 ⋅ 1.0 ⋅ 0.177 ⋅ 1.269 + 1.030 = 1.230 kN

Out-of-service wind loading per bayOut-of-service parallel wind on boards and transoms per bay

Fwo||b = qp(z = 13.00m) ⋅ cs||u ⋅ (cf||b ⋅ Wtotal ⋅ Lbay + cf,tr ⋅ Ltr ⋅ dt)= 0.888 ⋅ 1.0 ⋅ (0.02 ⋅ 1.637 ⋅ 2.00 + 0.8 ⋅ 5.511 ⋅ 0.048) = 0.247 kN

Out-of-service parallel wind on toe boards per bay

Fwo||tb = qp(z = 13.00m) ⋅ cs||u ⋅ cf||tb ⋅ Htb ⋅ Lbay = 0.888 ⋅ 1.0 ⋅ 0.1 ⋅ 0.225 ⋅ 2.00= 0.040 kN

Out-of-service parallel wind on brick guards per bay

Fwo||bg = qp(z = 13.00m) ⋅ cs||u ⋅ cf||bg ⋅ Lbay ⋅ 1.0 = 0.888 ⋅ 1.0 ⋅ 0.073 ⋅ 2.00 ⋅ 1.0= 0.130 kN

Out-of-service wind on transoms per unboarded lift bay

Fwo||tu = qp(z = 8.00m) ⋅ cs||u ⋅ cf,t ⋅ Ltr ⋅ dt = 0.838 ⋅ 1.0 ⋅ 1.2 ⋅ 5.511 ⋅ 0.048 = 0.268 kN

Out-of-service wind on unbraced intermediate frame tubes

Fwo||ui = qp(z = 13.00m) ⋅ cs||u ⋅ cf,t ⋅ Lt,ui ⋅ dt = 0.888 ⋅ 1.0 ⋅ 1.2 ⋅ 4.000 ⋅ 0.048= 0.206 kN

Out-of-service wind on unbraced intermediate frame tubes of unboarded lifts

Fwo||ui,u = qp(z = 8.00m) ⋅ cs||u ⋅ cf,t ⋅ Lt,ui ⋅ dt = 0.838 ⋅ 1.0 ⋅ 1.2 ⋅ 4.000 ⋅ 0.048= 0.194 kN

Out-of-service wind on ledger-braced intermediate frame tubes

Fwo||bi = qp(z = 13.00m) ⋅ cs||u ⋅ cf,t ⋅ Lt,bi ⋅ dt = 0.888 ⋅ 1.0 ⋅ 1.2 ⋅ 6.690 ⋅ 0.048= 0.344 kN

Out-of-service wind on ledger-braced intermediate frame tubes of unboarded lifts

Fwo||bi,u = qp(z = 8.00m) ⋅ cs||u ⋅ cf,t ⋅ Lt,bi ⋅ dt = 0.838 ⋅ 1.0 ⋅ 1.2 ⋅ 6.690 ⋅ 0.048= 0.325 kN

Out-of-service wind loading distribution to the windward end standardsOut-of-service wind on boarded lift outer end standards

Fwo||be,o = fo ⋅ Fwo||bgne = 0.266 ⋅ 1.230 = 0.327 kN




Out-of-service wind on boarded lift inner end standards

Fwo||be,i = fi ⋅ Fwo||bgne = 0.734 ⋅ 1.230 = 0.903 kN

Out-of-service wind on unboarded lift outer end standards

Fwo||ue,o = fo ⋅ Fwo||ue = 0.266 ⋅ 0.414 = 0.110 kN

Out-of-service wind on unboarded lift inner end standards

Fwo||ue,i = fi ⋅ Fwo||ue = 0.734 ⋅ 0.414 = 0.304 kN

Out-of-service wind loading distribution to the intermediate standardsOut-of-service wind on unbraced intermediate outer standards of boarded lifts

Fwo||ui,b,o = 0.5 ⋅ Fwo||ui + fo ⋅ Fwo||b + Fwo||tb + Fwo||bg= 0.5 ⋅ 0.206 + 0.266 ⋅ 0.247 + 0.040 + 0.130 = 0.338 kN

Out-of-service wind on unbraced intermediate inner standards of boarded lifts

Fwo||ui,b,i = 0.5 ⋅ Fwo||ui + fi ⋅ Fwo||b = 0.5 ⋅ 0.206 + 0.734 ⋅ 0.247 = 0.284 kN

Out-of-service wind on ledger-braced outer intermediate standards of boarded lifts

Fwo||bi,b,o = 0.5 ⋅ Fwo||bi + fo ⋅ Fwo||b + Fwo||tb + Fwo||bg= 0.5 ⋅ 0.344 + 0.266 ⋅ 0.247 + 0.040 + 0.130 = 0.407 kN

Out-of-service wind on ledger-braced inner intermediate standards of boarded lifts

Fwo||bi,b,i = 0.5 ⋅ Fwo||bi + fi ⋅ Fwo||b = 0.5 ⋅ 0.344 + 0.734 ⋅ 0.247 = 0.354 kN

Out-of-service wind on unbraced intermediate outer standards of unboarded lifts

Fwo||ui,u,o = 0.5 ⋅ Fwo||ui,u + fo ⋅ Fwo||tu = 0.5 ⋅ 0.194 + 0.266 ⋅ 0.268 = 0.168 kN

Out-of-service wind on unbraced intermediate inner standards of unboarded lifts

Fwo||ui,u,i = 0.5 ⋅ Fwo||ui,u + fi ⋅ Fwo||tu = 0.5 ⋅ 0.194 + 0.734 ⋅ 0.268 = 0.294 kN

Out-of-service wind on ledger-braced intermediate outer standards of unboarded lifts

Fwo||bi,u,o = 0.5 ⋅ Fwo||bi,u + fo ⋅ Fwo||tu = 0.5 ⋅ 0.325 + 0.266 ⋅ 0.268 = 0.233 kN

Out-of-service wind on ledger-braced intermediate inner standards of unboarded lifts

Fwo||bi,u,i = 0.5 ⋅ Fwo||bi,u + fi ⋅ Fwo||tu = 0.5 ⋅ 0.325 + 0.734 ⋅ 0.268 = 0.359 kN

Out-of-service wind distribution to the leeward end standardsOut-of-service wind on boarded lift outer leeward end standards

Fwo||be,l,o = fo ⋅ Fwo||bgne + fo ⋅ Fwo||b + Fwo||tb + Fwo||bg= 0.266 ⋅ 1.230 + 0.266 ⋅ 0.247 + 0.040 + 0.130 = 0.562 kN

Out-of-service wind on boarded lift inner leeward end standards

Fwo||be,l,i = fi ⋅ Fwo||bgne + fi ⋅ Fwo||b = 0.734 ⋅ 1.230 + 0.734 ⋅ 0.247 = 1.084 kN

Out-of-service wind on unboarded lift outer leeward end standards

Fwo||ue,l,o = fo ⋅ Fwo||ue + fo ⋅ Fwo||tu = 0.266 ⋅ 0.414 + 0.266 ⋅ 0.268 = 0.181 kN




Out-of-service wind on unboarded lift inner leeward end standards

Fwo||ue,l,i = fi ⋅ Fwo||ue + fi ⋅ Fwo||tu = 0.734 ⋅ 0.414 + 0.734 ⋅ 0.268 = 0.500 kN

Out-of-service wind loading on additional tubesOut-of-service wind on facade bracing Fwo||fb = qp(z = 13.00m) ⋅ cs||u ⋅ cf,t ⋅ dt ⋅

HliftLfb,1

= 0.888 ⋅ 1.0 ⋅ 1.2 ⋅ 0.048 ⋅2.00

2.828= 0.036 kN/m

Out-of-service wind per tie tube at the inner face

Fwo||tt,i = fi ⋅ qp(z = 13.00m) ⋅ cs||u ⋅ cf,t ⋅ dt ⋅ Ltt= 0.734 ⋅ 0.888 ⋅ 1.0 ⋅ 1.2 ⋅ 0.048 ⋅ 1.887 = 0.071 kN

Out-of-service wind per tie tube at the outer face

Fwo||tt,o = fo ⋅ qp(z = 13.00m) ⋅ cs||u ⋅ cf,t ⋅ dt ⋅ Ltt= 0.266 ⋅ 0.888 ⋅ 1.0 ⋅ 1.2 ⋅ 0.048 ⋅ 1.887 = 0.026 kN

Structural properties for 2D frame analysisTie support spring stiffness at the outer frame

ko = 10.4 kN/m

Tie support spring stiffness at the inner frame

ki = 54.3 kN/m

Type 4 steel tube section area Ag,t = 5.57 cm²

Type 4 steel tube moment of inertia

Ix,t = 13.77 cm⁴

Structural analysis

Analysis approachThe scaffold is idealised as two 2D vertical plane frames, one for the inner and one for the outer standards and ledgers. The dead and imposed loads from the boarding are apportioned to the inner and outer frames as noted above using simple statics. Self weights of tubes and components are also calculated and applied to the ledger and standard members. Wind forces acting in the direction parallel to the facade are also calculated as tabulated below.

The guard rails are not included in the stiffness analysis but their load reactions are applied to the standards as point loads. As an exception, if the scaffold is clad with sheeting or debris netting the guard rails at the top lift are included as members in the stiffness analysis.

In the frame analysis, the ledgers and standards are modelled as continuous at the internal joints but discontinuous (pinned) to the orthogonal members at the outer joints. For the relevant types of loads, this idealisation gives a satisfactory correlation with the 3D analysis developed for TG20:13 which allowed for the finite stiffness of right angle couplers.

The facade bracing members are considered to be pinned to the nodes at the ledger - standard intersections. The axial stiffness of the facade brace members is reduced substantially to allow for the actual stiffness of the end couplers and typical permissible eccentricity relative to the node points. The reduction factor was established by calibration of simplified models against accurate 3D analysis of braced bays and amounts to 1 / 75.

The foundations / base plates of the scaffold standards are modelled as pinned lift-off supports, which requires iterative analysis in some load combinations.

The effect of ties to the facade is modelled by means of equivalent spring supports acting horizontally. The stiffness values of these supports have been established by calibration against 3D analysis models in which the tie tubes are




connected to both the inner and outer ledgers with right angle couplers. The stiffness values depend on the width of the platform and whether the frame is the inner or outer relative to the facade.

The inner and outer 2D frames are subjected to a total of six load categories (in the positive and negative directions) applied horizontally in-plane:

1. Notional loads of 0.15 kN applied to the nodes at working lifts as required by BS EN 12811-1 clause 6.2.3.

2. Working wind loads applied to the members based on a peak velocity pressure of 0.2 kN/m².

3. Out-of-service wind loads applied to the members based on the peak velocity pressure calculated for the site exposure and the vertical position of the member.

All the above idealisations and approximations made to reduce complexity for 2D structural analysis have been validated against examples taken from the more accurate 3D analyses carried out during the development of TG20:13.

Vertical loads

Load description Inner face Outer face Unit

Dead load on unboarded lift ledgers 0.144 0.089 kN/m

Dead load on boarded lift ledgers 0.439 0.297 kN/m

Dead load per end standard per unboarded lift 0.251 0.269 kN

Dead load per end standard per boarded lift 0.424 0.411 kN

Dead load per unbraced intermediate standard per unboarded lift 0.098 0.196 kN

Dead load per ledger-braced intermediate standard per unboarded lift 0.169 0.267 kN

Dead load per unbraced intermediate standard per boarded lift 0.098 0.304 kN

Dead load per ledger-braced intermediate standard per boarded lift 0.169 0.375 kN

Dead load of facade bracing - 0.058 kN/m

Dead load per tie tube 0.072 0.034 kN

Imposed load on loaded lift ledgers 1.598 1.115 kN/m

Imposed load on 50% loaded lift ledgers 0.799 0.557 kN/m

Out-of-service imposed load on loaded lift ledgers 0.279 0.279 kN/m

Horizontal loads

In service Out of service UnitLoad description

Inner Outer Inner Outer

Notional horizontal load per working bay 0.150 0.150 - - kN

Wind on working lift end standards 0.232 0.084 0.903 0.327 kN

Wind on unboarded lift end standards 0.073 0.026 0.304 0.110 kN

Wind on unbraced intermediate standards of boarded lifts 0.064 0.076 0.284 0.338 kN

Wind on ledger-braced intermediate standards of boarded lifts 0.080 0.092 0.354 0.407 kN

Wind on unbraced intermediate standards of unboarded lifts 0.070 0.040 0.294 0.168 kN

Wind on ledger-braced intermediate standards of unboarded lifts 0.086 0.056 0.359 0.233 kN

Wind on working lift leeward end standards 0.272 0.137 1.084 0.562 kN




In service Out of service UnitLoad description

Inner Outer Inner Outer

Wind on unboarded lift leeward end standards 0.120 0.043 0.500 0.181 kN

Wind per tie tube 0.016 0.006 0.071 0.026 kN

Wind on facade bracing - 0.008 - 0.036 kN/m

Inner face Outer face

The analytical model is shown for the load combination which produces the maximum leg load: 5 - Dead + in-service imposed + wind (-ve).

For clarity, the point loads and horizontal loads are not labelled in the figures. Refer to the load tables above for values.

Analysis resultsMaximum leg load (kN)No. Load combination

Inner Outer

1 Dead + in-service imposed 9.3 7.6

2 Dead + in-service imposed + notional (+ve) 9.4 10.4

3 Dead + in-service imposed + notional (-ve) 9.4 10.9

4 Dead + in-service imposed + wind (+ve) 9.3 10.6

5 Dead + in-service imposed + wind (-ve) 9.3 11.0

6 Dead + out-of-service imposed 4.6 4.8

7 Dead + out-of-service imposed + wind (+ve) 4.7 9.1

8 Dead + out-of-service imposed + wind (-ve) 4.7 10.6

Maximum 9.4 11.0

Results summaryDescription Inner Outer




Description Inner Outer

Maximum unfactored leg load (kN) 9.4 11.0

TG20:13 checkCheck Result

TG20:13 compliance criteria check ✓ Pass

TG20:13 safe height check ✓ Pass

The scaffold height of 12.00 m is within the maximum safe height of a TG20:13 compliant tied independent scaffold of the given location and configuration.

ℹ This scaffold is TG20:13 compliant, with the following conditions in addition to the design criteria specified above:

No. TG20:13 compliance note

1 The maximum leg load should be communicated to the Engineer responsible for the scaffold foundation design.

2 Tie tubes should be connected to the inner and outer standards or ledgers with right-angle couplers.

Documents

Project no 0001 Date 11/09/2019 Notes Revision File Sample ... › wp-content › uploads › ...For tube and fitting scaffolding, in accordance with BS EN 12811-1:2003 and NASC TG20:13